2 Spontaneous Physical and Chemical Processes A waterfall runs downhillA lump of sugar dissolves in a cup of coffeeAt 1 atm, water freezes below 0 oC and ice melts above 0 oCHeat flows from a hotter object to a colder objectA gas expands in an evacuated bulbIron exposed to oxygen andwater forms rustspontaneousnonspontaneous
4 Does a decrease in enthalpy mean a reaction proceeds spontaneously? Spontaneous reactionsCH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = kJ/molH+ (aq) + OH- (aq) H2O (l) DH0 = kJ/molH2O (s) H2O (l) DH0 = 6.01 kJ/molNH4NO3 (s) NH4+(aq) + NO3- (aq) DH0 = 25 kJ/molH2O
5 Entropy (S) is a measure of the randomness or disorder of a system. DS = Sf - SiIf the change from initial to final results in an increase in randomnessSf > SiDS > 0For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state.Ssolid < Sliquid << SgasH2O (s) H2O (l)DS > 0
6 EntropyW = number of microstatesS = k ln WDS = Sf - SiDS = k lnWfWiWf > Wi then DS > 0Wf < Wi then DS < 0
7 Processes that lead to an increase in entropy (DS > 0)
9 EntropyState functions are properties that are determined by the state of the system, regardless of how that condition was achieved.Examples:energy, enthalpy, pressure, volume, temperature, entropyReviewPotential energy of hiker 1 and hiker 2 is the same even though they took different paths.
11 17.1Predict whether the entropy change is greater or less than zero for each of the following processes:(a) freezing ethanol(b) evaporating a beaker of liquid bromine at roomtemperature(c) dissolving glucose in water(d) cooling nitrogen gas from 80°C to 20°C
12 First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed.Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.Spontaneous process:DSuniv = DSsys + DSsurr > 0Equilibrium process:DSuniv = DSsys + DSsurr = 0
13 Entropy Changes in the System (DSsys) The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C.rxnaA + bB cC + dDDS0rxndS0(D)cS0(C)=[+]-bS0(B)aS0(A)DS0rxnnS0(products)=SmS0(reactants)-
14 17.2From the standard entropy values in Appendix 3, calculate the standard entropy changes for the following reactions at 25°C.CaCO3(s) CaO(s) + CO2(g)(b) N2(g) + 3H2(g) NH3(g)(c) H2(g) + Cl2(g) HCl(g)
15 Entropy Changes in the System (DSsys) When gases are produced (or consumed)If a reaction produces more gas molecules than it consumes, DS0 > 0.If the total number of gas molecules diminishes, DS0 < 0.If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number.
16 17.3Predict whether the entropy change of the system in each of the following reactions is positive or negative.2H2(g) + O2(g) 2H2O(l)(b) NH4Cl(s) NH3(g) + HCl(g)(c) H2(g) + Br2(g) 2HBr(g)
17 Entropy Changes in the Surroundings (DSsurr) Exothermic ProcessDSsurr > 0Endothermic ProcessDSsurr < 0
18 Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.S = k ln WW = 1S = 0
19 For a constant temperature and constant pressure process: Gibbs Free EnergySpontaneous process:DSuniv = DSsys + DSsurr > 0Equilibrium process:DSuniv = DSsys + DSsurr = 0For a constant temperature and constant pressure process:Gibbs free energy (G)DG = DHsys -TDSsysDG < The reaction is spontaneous in the forward direction.DG > The reaction is nonspontaneous as written. Thereaction is spontaneous in the reverse direction.DG = The reaction is at equilibrium.
20 DG0 of any element in its stable form is zero. The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.rxnaA + bB cC + dDDG0rxndDG0 (D)fcDG0 (C)=[+]-bDG0 (B)aDG0 (A)DG0rxnnDG0 (products)f=SmDG0 (reactants)-Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states.fDG0 of any element in its stable form is zero.f
24 Equilibrium Pressure of CO2 Temperature and Spontaneity of Chemical ReactionsCaCO3 (s) CaO (s) + CO2 (g)Equilibrium Pressure of CO2DH0 = kJ/molDS0 = J/K·molDG0 = DH0 – TDS0At 25 oC, DG0 = kJ/molDG0 = 0 at 835 oC
25 Gibbs Free Energy and Phase Transitions DG0 = 0= DH0 – TDS0H2O (l) H2O (g)DS =TDH=40.79 kJ/mol373 K= 1.09 x 10-1 kJ/K·mol= 109 J/K·mol
26 17.5 The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid → liquid and liquid → vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.
27 Gibbs Free Energy and Chemical Equilibrium DG = DG0 + RT lnQR is the gas constant (8.314 J/K•mol)T is the absolute temperature (K)Q is the reaction quotientAt EquilibriumDG = 0Q = K0 = DG0 + RT lnKDG0 = - RT lnK
28 Free Energy Versus Extent of Reaction DG0 > 0DG0 < 0
30 17.6Using data listed in Appendix 3, calculate the equilibrium constant (KP) for the following reaction at 25°C:2H2O(l) H2(g) + O2(g)
31 AgCl(s) Ag+(aq) + Cl-(aq) 17.7In Chapter 16 we discussed the solubility product of slightly soluble substances. Using the solubility product of silver chloride at 25°C (1.6 x 10-10), calculate ΔG° for the processAgCl(s) Ag+(aq) + Cl-(aq)
32 17.8 The equilibrium constant (KP) for the reaction N2O4(g) 2NO2(g) is at 298 K, which corresponds to a standard free-energy change of 5.40 kJ/mol. In a certain experiment, the initial pressures are PNO2 = atm and PN2O4 = atm.Calculate ΔG for the reaction at these pressures and predict the direction of the net reaction toward equilibrium.